Question #fb6ad

Question

1530 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T20:12:11

See Below

Right Side#=(sinx-cscx)/(cosx-cotx)#

#=(sinx-1/sinx)/(cosx-cosx/sinx)#

#=((sin^2x-1)/sinx) / ((cosx sinx-cosx)/sinx)#

#=(sin^2x-1)/sinx xx sinx/(cosx sinx-cosx)#

#=(sin^2x-1)/(cosx sinx-cosx)#

#=(-1(1-sin^2x))/(-cosx(1-sinx))#

#=cos^2x/(cosx(1-sinx))#

#=cosx/(1-sinx)#

#=#Left Side